Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14D Solutions
Exercise 14D
(Q1) Find the slope and y – intercept of the line:
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
(i) Y = 4
General equation, y = mx + c
Given equation is y = 4
Y = 0x + 4
m = slope
∴ m = 0
And y – intercept = 4
(ii) ax – by = 0
General equation, y = mx + c
We have to convert given equation in general equation.
ax – by = 0
ax = by
a/b x = y
y = a/b x + 0
∴ m = a/b and y intercept = 0
(iii) 3x – 4y = 5
General equation, y = mx + c
We have to convert given equation in general equation.
3x – 4y = 5
3x = 5 + 4y
3x – 5 = 4y
3x – 5/4 = y
Y = 3x/4 -5/4
∴ Slope (m) = ¾ and
y – Intercept = -5/4
(Q2) The equation of a line x – y = 4. Find its slope and y-intercept also find its inclination.
Solution:
Given equation of line is x – y = 4
We have to find slope (m) =? Y-Intercept =? Inclination (θ) = ?
General equation, y = mx + c
We have to convert given equation in general form,
x – y = 4
x – 4 = y
y = x – 4
∴ Slope (m) = 1 and
y – Intercept = -4
We know that,
m = tan θ
∴ tan θ = 1
tan θ = tan 45°
∴ θ = 45°
∴ Inclination (θ) = 45°
(Q3) (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4
Perpendicular to the line 3x – y =?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y =?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
If the two lines are perpendicular then the slope is-
Given lines are 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0
Now, we have to find slope of these lines –
First line => 3x + 4y + 7= 0
4y = -3x -7
Y = -3x -7/4
Y = (-3/4) x + (-7/4)
Comparing this equation with the general form of y = mx + c
∴ m1 = -3/4
Second line: 28x – 21y + 50 = 0
-21y = -28x – 50
y = -28x-50/-21
y = -28/-21 x + (-50/-20)
y = 28/21 x + (50/21)
y = (4/3) x + (50/21)
comparing this equation with general equation y = mx + c
∴ m2 = 4/3
We know that,
m1 × m2 = -1
(-3/4) × (4/3) = -1
-12/12 = -1
1 = -1
∴ Two lines are perpendicular
(ii) If the two lines are perpendicular then the slope is ‘-1’ that is,
i.e. m1 × m2 = 1
Given lines are x – 3y = 4 and 3x – y = 7
Now, we have to find slope of these lines-
First line => x – 3y = 4
-3y = 4 – x
Y = 4-x/-3
Y = 4/-3 + (-x/-3)
= x/3 + (4/-3)
Y = 1/3 x + (4/-3)
Comparing this equation with the general form of y = mx + c
∴ m1 =1/3
Second line => 3x – y = 7
-y = 7 – 3x
y = -7 + 3x
y = 3x – 7
y = 3x + (-7)
Comparing this equation with general form of y = mx + c
∴ m2 = 3
∴ We know that,
m1 × m2 = -1
3×1/3 ≠ -1
∴ the slopes of two lines are 1 not equal to -1 .
Hence the two lines are not perpendicular to each other.
(iii) If the two lines are parallel then the slope are same.
Given lines are 3x + 2y = 5 and x + 2y = 1
Now, we have to find slope of these lines-
First line=> 3x + 2y = 5
2y = 5-3x
2y = (-3)x + 5
Y = (-3)x + 5/2
Y = (-3/2)x + 5/2
Comparing this equation with the general form of y = mx + c
∴ m1 = -3/2
∴ Second line => x + 2y = 1
2y = 1-x
2y = -x +1
y = -x+1/2
y = (-1/2)x + ½
Comparing this equation with the general form of y = mx + c
∴ m2 = -1/2
We can see that the slope of m1 and m2 are not equal or same.
Hence, the two lines are not parallel to each other
(iv) Given that, slope (m) = 2
We know that,
y2 – y1 /x2 – x1 = m
(1, 4) = (x1, y1) and (x, 2) = (x2, y2)
2-4/x-1 = 2
-2 = 2× (x -1)
-2 = 2x -2
-2+2 = 2x
0 = 2x
x = 0
(Q4) Find the slope of the line which is parallel to:
(i) x + 2y + 3 = 0
(ii) x/2 – y/3 – 1= 0
Solution:
Given equation of line is x+2y + 3 = 0
First we have to convert given equation in general form y = mx + c.
X + 2y + 3 = 0
2y = -x-3
y = -x-3/2
y = -x/2 + (-3/2)
y = -1/2 x + (-3/2)
Comparing this equation with y = mx + c
∴ m = -1/2
(ii) Given equation of line is x/2 – y/3 – 1 = 0
First we have to convert given equation with y = mx + c
x/3 – y/3 -1 = 0
-y/3 = -x/3 +1
Multiply by ‘3’ on both sides –
3 × (-y/3) = 3 × (-x/3) + 3
-y = -3x + 3
y = 3x – 3
Comparing this equation with y = mx + c
∴ Slope (m) = 3
(Q5) Find the slope of the line which is perpendicular to:
(i) x – y/2 + 3 = 0
(ii) x/3 – 2y = 4
Solution:
Given equation of line is
x – y/2 + 3= 0
First we have to convert given equation in general form of y = mx + c
x –y/2 + 3 = 0
-y/2 = -3 – x
-y = 2(-3 –x)
-y = – 6 – 2x
y = 6 + 2x
y = 2x + 6
Comparing with y = mx +c
∴ Slope (m1) = 2
The slope of lines is perpendicular is –
m1 × m2 =- -1
m2 = -1/m1
m2 = -1/2
Slope (m2) = -1/2
(ii) Given equation of line is
x/3 – 2y = 4
First we have to convert given equation in general form of y = mx + c
x/3 – 2y = 4
-2y = 4 – x/3
y = -2 (4- x/3)
= -8 + 2x/3
y = 2/3 x + (-8)
Comparing this equation with y = mx + c
∴ Slope (m1) = 2/3
The slope of line is perpendicular to the line-
m1 × m2 = -1
m2 = -1/m1
m2 = -1/(2/3)
m2 = -3/2
(Q6) Find the value of P if the lines, whose equations are 2x – y + 5 and px + 3y = 4 are perpendicular to each other.
Solution:
Given equations are –
2x – y + 5 = 0 —– (1) and
Px + 3y = 4 —– (2)
First we have to convert equation (1) into general form of y = mx + c
2x – y + 5 = 0
-y = – 2x – 5
y = 2x +5
Comparing this equation with y = mx + c
∴ Slope (m1) = 2
Now, we have to convert equation (2) in general form of y = mx + c
Px + 3y = 4
3y = 4 – px
3y = – px + 4
y = – px + 4/3
y = -p/3 x + 4/3
Comparing this equation with y = mx + c
∴ Slope (m2) = -p /3
If two lines are perpendicular then the product of slopes of two line is -1
m1 × m2 = -1
2 × -p/3 = -1
-2p/3 = -1
2p/3 = 1
P = 3/2
(Q7) The equation of a line AB is 2x – 2y + 3 = 0
(i) Find the slope of the line AB
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
Given equation of line is 2x – 2y + 3 = 0
(i) First we have to find slope of this line.
So, now we have to convert given equation of line into the general form of y = mx + c
2x – 2y + 3 = 0
-2y = -2x -3
2y = 2x +3
y = 2x + 3/2
y = 2x/2 + 3/2
y = x + 3/2
Comparing this equation with y = mx + c
Slope (m) = 1
∴ Slope of line AB = 1
(ii) The slope of line AB is 1
So, we know that the value of tan 45° = 1
∴ We know that,
Slope = tan θ
= 1
And tan 45° = 1
∴ θ = 45°
(Q8) The lines represented by 4x +3y = 9 and px – 6y + 3 = 0 are parallel find the value of p.
Solution:
Given equation of lines are 4x + 3y = 9 —– (1)
And px – 6y + 3 = 0
First we have to convert equation (1) in general form of y = mx + c
4x + 3y = 9
3y = 9 – 4x
3y = – 4x + 9
y = – 4x+9/3
y = -4x/3 + 9/3
y = -4/3 x + 9/3
∴ Slope (m1) = -4/3
Now, we have to convert equation (2) in general form of y = mx + c
Px – 6y + 3 = 0
-6y = -3 – px
-6y = – px -3
y = px-3/ -6
y = -px/-6 + (-3/-6)
= px/6 + 3/6
Y = px/6 + ½
∴ Slope (m2) = p/6
Also given that two lines are parallel.
We know that, if two line are parallel then the slopes of two lines are same or equal.
∴ m1 = m2
-4/3 = p/6
-4 = 3p/6
-4 = p/2
-8 = p
P = -8
(Q9) If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution:
Given lines are y = 3x + 7 —— (1)
And 2y + px = 3 —— (2)
First we have to find slope
y = 3x + 7
Comparing this equation with y = mx + c
∴ Slope (m1) = 3
Now, we have to convert equation (2) in general form of y = mx + c
2y + px = 3
2y = 3 – px
2y = – px + 3
y = – px + 3/2
y = -p/2 x +3/2
∴ Slope m2 = -p/2
Also, given that, two lines are perpendicular
We know that,
If two lines are perpendicular then the slopes of two lines are – 1.
m1 × m2 = -1
3 × -p/2 = -1
-3p/2 = -1
P = 2/3
Here is your solution of Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14D
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